A132276 Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,k), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0) (0<=k<=n).

1, 1, 1, 3, 2, 1, 6, 7, 3, 1, 16, 18, 12, 4, 1, 40, 53, 37, 18, 5, 1, 109, 148, 120, 64, 25, 6, 1, 297, 430, 369, 227, 100, 33, 7, 1, 836, 1244, 1146, 760, 385, 146, 42, 8, 1, 2377, 3656, 3519, 2518, 1391, 606, 203, 52, 9, 1, 6869, 10796, 10839, 8188, 4900, 2346, 903, 272

Offset

0,4

Comments

Mirror image of A059397. - Emeric Deutsch, Aug 18 2007

Row sums yield A059398.

Riordan matrix (g(x),x*g(x)), where g(x) = (1-x-x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^2). - Emanuele Munarini, May 05 2011

References

Lin Yang and S.-L. Yang, The parametric Pascal rhombus. Fib. Q., 57:4 (2019), 337-346.

Links

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Paul Barry, On Motzkin-Schröder Paths, Riordan Arrays, and Somos-4 Sequences, J. Int. Seq. (2023) Vol. 26, Art. 23.4.7.

W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.

Sheng-Liang Yang and Yuan-Yuan Gao, The Pascal rhombus and Riordan arrays, Fib. Q., 56:4 (2018), 337-347. See Fig. 1.

Formula

T(n,0) = A128720(n).

G.f.: G(t,z) = g/(1-t*z*g), where g = 1 +z*g +z^2*g +z^2*g^2 or g = c(z^2/(1-z-z^2)^2)/(1-z-z^2), where c = ((1-sqrt(1-4*z))/(2*z) is the Catalan function.

T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) + T(n-2,k). - Emeric Deutsch, Aug 18 2007

Column k has g.f. z^k*g^(k+1), where g = 1 +z*g +z^2*g +z^2*g^2 = (1 -z-z^2 -sqrt((1+z-z^2)*(1-3*z-z^2)))/(2*z^2).

T(n,k) = Sum_{i=0..(n-k)/2} (binomial(2*i+k,i)*(k+1)/(i+k+1)* Sum_{j=0..(n-k-2*i)} binomial(i+j+k,i+k)*binomial(j,n-k-2*i-j). - Emanuele Munarini, May 05 2011

Example

T(3,2) = 3 because we have UUh, UhU and hUU.
Triangle begins:
    1;
    1,   1;
    3,   2,   1;
    6,   7,   3,  1;
   16,  18,  12,  4,  1;
   40,  53,  37, 18,  5, 1;
  109, 148, 120, 64, 25, 6, 1;
  ...

Maple

g:=((1-z-z^2-sqrt((1+z-z^2)*(1-3*z-z^2)))*1/2)/z^2: G:=simplify(g/(1-t*z*g)): Gser:=simplify(series(G, z=0, 13)): for n from 0 to 10 do P[n]:=sort(coeff(Gser, z, n)) end do: for n from 0 to 10 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form

Mathematica

Flatten[Table[Sum[Binomial[2i+k, i(k+1)/(i+k+1)*Sum[Binomial[i+j+k, i+k]* Binomial[j, n-k-2i-j], {j, 0, n-k-2i}], {i, 0, (n-k)/2}], {n, 0, 15}, {k, 0, n}]] (* Emanuele Munarini, May 05 2011 *)
c[x_] := (1 - Sqrt[1 - 4*x])/(2*x); g[z_] := c[z^2/(1 - z - z^2)^2]/(1 - z - z^2); G[t_, z_] := g[z]/(1 - t*z*g[z]); CoefficientList[ CoefficientList[Series[G[t, x], {x, 0, 49}, {t, 0, 49}], x], t]//Flatten (* G. C. Greubel, Dec 02 2017 *)

Prog

(Maxima) create_list(sum(binomial(2*i+k, i) * (k+1)/(i+k+1) * sum(binomial(i+j+k, i+k) * binomial(j, n-k-2*i-j), j, 0, n-k-2*i), i, 0, (n-k)/2), n, 0, 15, k, 0, n); /* Emanuele Munarini, May 05 2011 */
(PARI) for(n=0, 10, for(k=0, n, print1(sum(i=0, (n-k)/2, (binomial(2*i+k, i) *(k+1)/(i+k+1)*sum(j=0, (n-k-2*i), binomial(i+j+k, i+k)*binomial(j, n-k-2*i-j)))), ", "))) \\ G. C. Greubel, Nov 29 2017

Crossrefs

Cf. A059397, A128720 (the leading diagonal).

Cf. A059398.

Sequence in context: A267121 A208518 A139624 * A257558 A370470 A202390

Adjacent sequences: A132273 A132274 A132275 * A132277 A132278 A132279

Keyword

nonn,tabl

Author

Emeric Deutsch, Aug 16 2007, Sep 03 2007

Status

approved